A 6.0-cm-diameter cylinder of nitrogen gas has a 4.0-cm-thick movable copper piston. The cylinder is oriented vertically, as shown in FIGURE P19.49, and the air above the piston is evacuated. When the gas temperature is 20°C, the piston floats 20 cm above the bottom of the cylinder.

c. What is the new equilibrium temperature of the gas?

Question

A 6.0-cm-diameter cylinder of nitrogen gas has a 4.0-cm-thick movable copper piston. The cylinder is oriented vertically, as shown in FIGURE P19.49, and the air above the piston is evacuated. When the gas temperature is 20°C, the piston floats 20 cm above the bottom of the cylinder.

c. What is the new equilibrium temperature of the gas?

Accepted Answer

Hey everyone. So this problem is dealing with the ideal gas law. Let's see what it's asking us. A cube shaped container with a side length of eight centimeters as a movable piston that contains carbon dioxide gas at a temperature of 27 °C. The movable piston is made of steel and is 1.5 centimeters thick at 27 °C. The piston is found to have a position five centimeters above the bottom of the cube. The space above the pit, excuse me, the space above the piston is evacuated. What is the new equilibrium temperature of the gas? When 3.5 joules of heat energy has been transferred into the gas? Assume the density of steel is 8000 kg per cubic meter. There are multiple choice answers in units of Kelvin are a 971 B 921 C 725 D 473 or E 298. OK. So this problem is asking us for that new equilibrium temperature. So T two, which will be the initial temperature plus the change in temperature or T one plus delta T, we are also told about the amount of heat energy that is introduced into the gas. And so we can recall the equation Q or heat energy is equal to N multiplied by C sub P multiplied by delta T. So delta T is going to be our target for this equation or for this problem. And then cap our molar heat capacity for carbon dioxide which is a constant. So, and we have our heat energy or Q. And so to solve for delta T, we still need to find N or the moles of gas in this problem. And so we can recall the equation PV equals N RT which is our ideal gas law. And so isolating N on the left hand side of the equation gives us P multiplied by B divided by R multiplied by T R is the ideal gas constant. And we can solve for our pressure to solve for N thinking about the piston as a singular point, we can draw our free body diagram where we have the force from the gas acting in the positive Y direction. And then the weight of the piston acting in the negative Y direction because we are told that we are at the new equilibrium temperature, we can say that the sum of the forces in the Y direction is zero, which means that the force of the gas is equal to our weight. The force of the gas can be rewritten as the pressure multiplied by the area because we can recall that pressure by definition is force per area. And then the weight can be rewritten as mass multiplied by gravity. We don't know the mass of the piston, but we do know the density and the material. And so we have the density multiplied by the volume is equal to the mass multiplied by gravity. In turn volume can be expanded as the area multiplied by the thickness. And so we can cancel the areas on both sides of that equation. And now we have pressures equal to row multiplied by T multiplied by G. So we plug that in to our equation for the moles of gas, we have row multiplied by G multiplied by the thickness of the piston. And then this is the volume of the gas. So that will be the length multiplied by the width multiplied by the height of this cube container. And then all of that divided by our ideal gas constant multiplied by our initial temperature. And so when we go to plug in those known values, we'll have 8000 kg per meter cubed multiplied by gravity 9.8 m per second squared multiplied by the thickness of the piston which was 1.5 centimeters, keeping everything in standard units, that's 0.015 m. And then we have this cube shaped container that's eight centimeters square. So that will be 0.8 m squared multiplied by the height which was at 0.05 m or five centimeters when it was at 27 °C. Our ideal gas constant we can recall is 8.1 34 jules per Kelvin multiplied by that initial temperature 300 degrees Kelvin when we convert from Celsius to Kelvin. And so that gives us an N of 1.51 times 10 to the negative four moles. Now, we can simply plug that into our delta T equation. So we have delta T is equal to Q divided by N multiplied by C sub P where Q was 3.5 joules. And we just saw for one point 51 times 10 to the negative four moles. And then CCP four carbon dioxide is 37.35 joules per mole. Kelvin such as a constant, you can look up and that comes out to 621 degrees Kelvin. And so our final temperature T two is equal to our initial temperature of 300 degrees Kelvin plus our delta T 621 degrees Kelvin, which gives us 921 degrees Kelvin. And so that's the final answer for this problem. And we can see that, that aligns with answer choice. A so that's all we have for this one. We'll see you in the next video.

Verified Solution

Key Concepts

Ideal Gas Law

Thermal Equilibrium

Pressure-Volume Relationship